Introduction to Open Data Science 2018 is a course about modern methods and tools in data science.
You can find my course github repository here https://github.com/eavalo/IODS-project/.
Read in the dataset. The dataset contains 7 variables:
learning2014 <- read.csv("~/git/IODS-project/data/learning2014.csv")
Check that the data was read correctly. Print out the first few rows and check the type of the columns:
head(learning2014)
## gender age attitude deep surf stra points
## 1 F 53 37 3.583333 2.583333 3.375 25
## 2 M 55 31 2.916667 3.166667 2.750 12
## 3 F 49 25 3.500000 2.250000 3.625 24
## 4 M 53 35 3.500000 2.250000 3.125 10
## 5 M 49 37 3.666667 2.833333 3.625 22
## 6 F 38 38 4.750000 2.416667 3.625 21
str(learning2014)
## 'data.frame': 166 obs. of 7 variables:
## $ gender : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: int 37 31 25 35 37 38 35 29 38 21 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...
Explore the data by plotting
library(GGally)
library(ggplot2)
# Define the plot.
p <- ggpairs(learning2014, mapping = aes(col = gender, alpha=0.3),
lower = list(combo = wrap("facethist", bins = 20)), legend=1)
# Draw the plot
p
There are almost twice as many women in the dataset compared to men. The distribution of age is skewed towards higer values. The distribution of attitutde, deep, surf, stra and points seems to be more normally distributed. The variable with the highest correlation with points is attitude.
summary(learning2014$gender)
## F M
## 110 56
summary(learning2014$age)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 17.00 21.00 22.00 25.51 27.00 55.00
summary(learning2014$attitude)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 14.00 26.00 32.00 31.43 37.00 50.00
summary(learning2014$deep)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.583 3.333 3.667 3.680 4.083 4.917
summary(learning2014$surf)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.583 2.417 2.833 2.787 3.167 4.333
summary(learning2014$stra)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.250 2.625 3.188 3.121 3.625 5.000
summary(learning2014$points)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 7.00 19.00 23.00 22.72 27.75 33.00
Fit a regression model using points as the outcome variable and age, gender and attitude as explanatory variables:
# Fit the linear regression model
lm_model <- lm(points ~ age + gender + attitude, data=learning2014)
# Sumamry of the model
summary(lm_model)
##
## Call:
## lm(formula = points ~ age + gender + attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.4590 -3.3221 0.2186 4.0247 10.4632
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.42910 2.29043 5.863 2.48e-08 ***
## age -0.07586 0.05367 -1.414 0.159
## genderM -0.33054 0.91934 -0.360 0.720
## attitude 0.36066 0.05932 6.080 8.34e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.315 on 162 degrees of freedom
## Multiple R-squared: 0.2018, Adjusted R-squared: 0.187
## F-statistic: 13.65 on 3 and 162 DF, p-value: 5.536e-08
Gender and age are not significantly associated to the points so remove them from the model. Fit the model again using only attitude as the explanatory variable.
lm_model_2 <- lm(points ~ attitude, data=learning2014)
summary(lm_model_2)
##
## Call:
## lm(formula = points ~ attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.63715 1.83035 6.358 1.95e-09 ***
## attitude 0.35255 0.05674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09
Explanatory variable attitude is statistically significantly associated to the outcome variable points with a p-value of 4.12e-09. The estimate of the regression coefficient is 0.35 meaning a one unit increase in attitude is on average associated to a 1.42 unit increase in points. The R² of the model is 0.19 which means that the variation of the explanatory variable attitude explains 19% of the variation of the outcome variable points.
Plot the residuals versus the fitted values
plot(lm_model_2, which=1)
Plot the normal QQ-plot:
plot(lm_model_2, which=2)
Plot the residuals versus leverage plot:
plot(lm_model_2, which=5)
The dataset has been constructed by joining two student alchol comsuption datasets which were downloaded from https://archive.ics.uci.edu/ml/datasets/Student+Performance. The two datasets contain the same variables and the students are partially overlapping. The data was joined by using following columns as surrogate identifiers for students:
The variables not used for joining have been combined by averaging for numeric columns and by taking the first answer for non-numeric columns. Two new variables have been define:
Read in the dataset.
alc <- read.csv("~/git/IODS-project/data/alc.csv")
Variables in the data set:
str(alc)
## 'data.frame': 382 obs. of 35 variables:
## $ school : Factor w/ 2 levels "GP","MS": 1 1 1 1 1 1 1 1 1 1 ...
## $ sex : Factor w/ 2 levels "F","M": 1 1 1 1 1 2 2 1 2 2 ...
## $ age : int 18 17 15 15 16 16 16 17 15 15 ...
## $ address : Factor w/ 2 levels "R","U": 2 2 2 2 2 2 2 2 2 2 ...
## $ famsize : Factor w/ 2 levels "GT3","LE3": 1 1 2 1 1 2 2 1 2 1 ...
## $ Pstatus : Factor w/ 2 levels "A","T": 1 2 2 2 2 2 2 1 1 2 ...
## $ Medu : int 4 1 1 4 3 4 2 4 3 3 ...
## $ Fedu : int 4 1 1 2 3 3 2 4 2 4 ...
## $ Mjob : Factor w/ 5 levels "at_home","health",..: 1 1 1 2 3 4 3 3 4 3 ...
## $ Fjob : Factor w/ 5 levels "at_home","health",..: 5 3 3 4 3 3 3 5 3 3 ...
## $ reason : Factor w/ 4 levels "course","home",..: 1 1 3 2 2 4 2 2 2 2 ...
## $ nursery : Factor w/ 2 levels "no","yes": 2 1 2 2 2 2 2 2 2 2 ...
## $ internet : Factor w/ 2 levels "no","yes": 1 2 2 2 1 2 2 1 2 2 ...
## $ guardian : Factor w/ 3 levels "father","mother",..: 2 1 2 2 1 2 2 2 2 2 ...
## $ traveltime: int 2 1 1 1 1 1 1 2 1 1 ...
## $ studytime : int 2 2 2 3 2 2 2 2 2 2 ...
## $ failures : int 0 0 2 0 0 0 0 0 0 0 ...
## $ schoolsup : Factor w/ 2 levels "no","yes": 2 1 2 1 1 1 1 2 1 1 ...
## $ famsup : Factor w/ 2 levels "no","yes": 1 2 1 2 2 2 1 2 2 2 ...
## $ paid : Factor w/ 2 levels "no","yes": 1 1 2 2 2 2 1 1 2 2 ...
## $ activities: Factor w/ 2 levels "no","yes": 1 1 1 2 1 2 1 1 1 2 ...
## $ higher : Factor w/ 2 levels "no","yes": 2 2 2 2 2 2 2 2 2 2 ...
## $ romantic : Factor w/ 2 levels "no","yes": 1 1 1 2 1 1 1 1 1 1 ...
## $ famrel : int 4 5 4 3 4 5 4 4 4 5 ...
## $ freetime : int 3 3 3 2 3 4 4 1 2 5 ...
## $ goout : int 4 3 2 2 2 2 4 4 2 1 ...
## $ Dalc : int 1 1 2 1 1 1 1 1 1 1 ...
## $ Walc : int 1 1 3 1 2 2 1 1 1 1 ...
## $ health : int 3 3 3 5 5 5 3 1 1 5 ...
## $ absences : int 5 3 8 1 2 8 0 4 0 0 ...
## $ G1 : int 2 7 10 14 8 14 12 8 16 13 ...
## $ G2 : int 8 8 10 14 12 14 12 9 17 14 ...
## $ G3 : int 8 8 11 14 12 14 12 10 18 14 ...
## $ alc_use : num 1 1 2.5 1 1.5 1.5 1 1 1 1 ...
## $ high_use : logi FALSE FALSE TRUE FALSE FALSE FALSE ...
I choose the following 4 variables for studying their relationship to high/low alcohol consumption:
I hypotize that high amount of “going out with friends” would be associated with a higer alcohol consumption since alcohol is often consumed in social situations. I think male gender could be associated to higher alcohol consumption just because males can tolerate more alcohol. I hypotize that higher “weekly study time” is associated to low alcohol compsuption: students with high study times are focused on school and don’t have as much time to drink alcohol. I also hypotize that individuals “with a romatic relationship” will consume less aclhol since they spend more time with their partners than friens and alchol is more often spend with friends.
Explore the variables of interest in regard to alchohol use
library(tidyr)
library(dplyr)
library(ggplot2)
# Explore the mean 'goout' and 'studytime' with respect to high/low alcohol use
alc %>% group_by(high_use) %>% summarise(count = n(), mean_goout=mean(goout),
mean_studytime=mean(studytime))
## # A tibble: 2 x 4
## high_use count mean_goout mean_studytime
## <lgl> <int> <dbl> <dbl>
## 1 FALSE 268 2.85 2.15
## 2 TRUE 114 3.72 1.77
# Explore high/low alcohol use stratified by 'sex'
alc %>% group_by(high_use, sex) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups: high_use [?]
## high_use sex count
## <lgl> <fct> <int>
## 1 FALSE F 156
## 2 FALSE M 112
## 3 TRUE F 42
## 4 TRUE M 72
# Explore high/low alcohol use stratified by 'romantic'
alc %>% group_by(high_use, romantic) %>% summarise(count = n())
## # A tibble: 4 x 3
## # Groups: high_use [?]
## high_use romantic count
## <lgl> <fct> <int>
## 1 FALSE no 180
## 2 FALSE yes 88
## 3 TRUE no 81
## 4 TRUE yes 33
# Draw barplots of variables of interest
###########################################
g_goout <- ggplot(alc, aes(x = goout, fill=high_use)) +
geom_bar() + xlab("Going out with friends") +
ggtitle("Going out with friends from 1 (very low) to 5 (very high) by alcohol use")
g_studytime <- ggplot(alc, aes(x = studytime, fill=high_use)) +
geom_bar() + xlab("Weekly study time") +
ggtitle("Weekly study time [1 (<2 hours), 2 (2 to 5 hours), 3 (5 to 10 hours), or 4 (>10 hours)] by alchol use")
g_sex <- ggplot(alc, aes(x = sex, fill=high_use)) +
geom_bar() +
ggtitle("Sex by alcohol use")
g_romantic <- ggplot(alc, aes(x = romantic, fill=high_use)) +
geom_bar() +
ggtitle("With a romantic relationship (yes/no) by alcohol use")
# Arrange the plots into a grid
library("gridExtra")
grid.arrange(g_goout, g_studytime, g_sex, g_romantic, ncol=2, nrow=2)
Based on the plots my assumptions seem to be somewhat corret:
Fit a logistic regression model using high_use as the targe variable and goout, studytime, sex and romantic as the explanatory variables.
# Fit the logistic regression model
m <- glm(high_use ~ goout + studytime + sex + romantic, data = alc, family = "binomial")
Summary of the fitted logistics regression model shows that goout, studytime and sex are statistically significantly associated to alchol comsumption. High alcohol comsumption is associated to high goout and male gender and low alcohol comsumption is associated to high studytime.
# Summary of the model
summary(m)
##
## Call:
## glm(formula = high_use ~ goout + studytime + sex + romantic,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.7365 -0.8114 -0.5009 0.9081 2.6642
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.6988 0.5712 -4.725 2.30e-06 ***
## goout 0.7536 0.1187 6.350 2.15e-10 ***
## studytime -0.4774 0.1683 -2.837 0.00456 **
## sexM 0.6657 0.2585 2.576 0.01000 *
## romanticyes -0.1424 0.2699 -0.528 0.59767
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 393.67 on 377 degrees of freedom
## AIC: 403.67
##
## Number of Fisher Scoring iterations: 4
Coefficients of the model as odds ratios and their confidence intervals:
# Calculate the odds ratios and confidence intervals of the coefficients
or <- coef(m) %>% exp
ci <- confint(m) %>% exp
# Print out the odds ratios and confidence intervals
cbind(or, ci)
## or 2.5 % 97.5 %
## (Intercept) 0.06728696 0.02129867 0.2010636
## goout 2.12456419 1.69404697 2.7003422
## studytime 0.62040325 0.44145946 0.8558631
## sexM 1.94589655 1.17538595 3.2443631
## romanticyes 0.86724548 0.50714091 1.4648961
From the odds ratios we can see that one unit increase in goout is associated with 2.1 times higher likelihood of high alchohol comsumption and one unit increase in studytime is associated with 0.6 times lower likelihood of high alcohol comsumption. Male gender is associated with 1.9 higer likelihood of high alcohol comsumption comapred to female gender. Being in a romatic relationship is not significantly associated to high/low alcohol comsuption in this model since the confidence intervals include 1.
My previously stated hypothesis seem to be verified by this model except that romantic is not associated to high or low alcohol comsumption in this model.
Fit a logistic model with the explanatory variables that were statistically significantly associated to high or low alchohol consumption:
# Fit the logistic regression model
m <- glm(high_use ~ goout + studytime + sex, data = alc, family = "binomial")
Prediction performance of the model
# Calcualte the predicted probabilities of high alcohol comspumtion
probability <- predict(m, type="response")
alc <- mutate(alc, probability=probability)
# Predict the high alcohol use with the probabilities
alc <- mutate(alc, prediction=probability > 0.5)
# Cross-tabulate the actual class and the predicted class
table(high_use = alc$high_use, prediction = alc$prediction)
## prediction
## high_use FALSE TRUE
## FALSE 250 18
## TRUE 76 38
The model seems to be quite good at predicting low alcohol use but performes less well in predicting high alcohol use.
Visualize the actual class, the predicted probabilities and the predicted class.
# Initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(alc, aes(x = probability, y = high_use, col=prediction))
# define the geom as points and draw the plot
g + geom_point()
Calculate the total proportion of mis-classified individuals using the regression model and with a simple guessing strategy where everyone is classified to be in the most prevalent class low use of alcohol.
# Define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# Call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2460733
# Compare the results to guessing that everybody belongs to the class low use of alcohol
loss_func(class = alc$high_use, prob = 0)
## [1] 0.2984293
Using the regression model 24.6% of the individuals are mis-classified compared to 29.8 % of mis-classified individuals when simply guessing everybody belongs to the low use of alcohol class. The model seems to provide some improvement to the simple guess of the most prevalent class.
Performe 10-fold cross-validation of the model to estiamte the performance of the model on unseen data. The performance of the model is measured with proportion of mis-classified individuals. The mean prediction error in the test set:
library(boot)
# 10-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# Mean prediction error
cv$delta[1]
## [1] 0.2382199
The mean prediction error in the test set is 0.25 which is better than the performance of the model introduced in the DataCamp exercises which had a mean prediction error of 0.26 in the test set.
Construct models with different number of predictors and calculate the test set and training set prediction errors.
# All the possible predictors
predictors <- c('school', 'sex', 'age', 'address', 'famsize', 'Pstatus', 'Medu',
'Fedu', 'Mjob', 'Fjob', 'reason', 'nursery', 'internet', 'guardian',
'traveltime', 'studytime', 'failures', 'schoolsup', 'famsup', 'paid',
'activities', 'higher', 'romantic', 'famrel', 'freetime', 'goout',
'health', 'absences', 'G1', 'G2', 'G3')
# Fit several models and record the test and traingin errors
# 1) Use all of the predictors.
# 2) Drop one predictor and fit a new model.
# 3) Continue until only one predictor is left in the model.
# Fit the models and calculate the erros
test_error <- numeric(length(predictors))
training_error <- numeric(length(predictors))
for(i in length(predictors):1) {
model_formula <- paste0("high_use ~ ", paste(predictors[1:i], collapse = " + "))
glmfit <- glm(model_formula, data = alc, family = "binomial")
# 10-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# Mean prediction error
test_error[i] <- cv$delta[1]
# Training error
training_error[i] <-
loss_func(alc$high_use, predict(glmfit,type="response"))
}
# Construct a table of prediction errors for plotting
data_error <- rbind(data.frame(n_predictors=1:length(predictors),
prediction_error=test_error,
type = "test error"),
data.frame(n_predictors=1:length(predictors),
prediction_error=training_error,
type = "training error"))
# Plot the test and training errors vs. number of predictors in the model
g <- ggplot(data_error, aes(x = n_predictors, y = prediction_error, col=type))
# define the geom as points and draw the plot
g + geom_point()
# Load libraries
library(corrplot)
library(dplyr)
Load in the Boston dataset from the MASS package.
library(MASS)
data("Boston")
str(Boston)
## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
The dataset has 14 variables and 506 observations. The following variables are present:
More details of the dataset can be found here https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html.
Summary of the variabes in the dataset:
summary(Boston)
## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00
Explore the distribution of the variables by plotting:
library(GGally)
library(ggplot2)
# Define the plot.
p <- ggpairs(Boston, mapping = aes(alpha=0.3),
lower = list(combo = wrap("facethist", bins = 20)))
# Draw the plot
p
Correlation of the variables:
cor(Boston) %>% corrplot(method="circle", type="upper", cl.pos="b", tl.pos="d")
Scale the dataset so that the mean of each variable is zero and standard deviation is one:
\[x_{scaled}=\frac{x - \mu_{x}}{\sigma_{x}}\],
where \(\mu_{x}\) is the mean of x and \(\sigma_{x}\) the standard deviation of x.
boston_scaled <- scale(Boston) %>% as.data.frame()
summary(boston_scaled)
## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865
From the summary we can see that the mean of the scaled variables is zero.
Create a factor variable críme from the crim (per capita crime rate by town) by dividing the crim variable by quartiles to ‘low’, ‘med_low’, ‘med_high’ and ‘high’ categories:
# Create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
# Create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE,
label=c("low", "med_low", "med_high", "high"))
# Remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# Add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
Divide the dataset to training and test sets so that 80% belongs to the training set and 20% belongs to the test set.
# Set seed so the results are reproducible
set.seed(1234)
# Take randomly 80% of the observations to the training set
train.idx <- sample(nrow(boston_scaled), size = 0.8 * nrow(boston_scaled))
train <- boston_scaled[train.idx,]
# Take the remaining 20% to the test set
test <- boston_scaled[-train.idx,]
Fit the linear discriminant analysis (LDA) on the training set using the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
The LDA biplot:
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col=classes, pch=classes)
lda.arrows(lda.fit, myscale = 2)
Use the fitted LDA model to predict the categorical crime rate in the test set. Cross tabulate the observed classes and the predicted classes in the test set:
# Save the correct classes from test data
correct_classes <- test$crime
# Remove the crime variable from test data
test <- dplyr::select(test, -crime)
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
## predicted
## correct low med_low med_high high
## low 11 8 0 0
## med_low 7 19 2 0
## med_high 1 6 18 1
## high 0 0 0 29
Model seems to perform perfectly at predicting the ‘high’ class and also predicts the other classes reasonably well. The prediction accuracy is worst for the ‘low’ class. The model mis-classifies a big proportion of the ‘low’ observations as ‘med_low’.
Reload the Boston dataset and standardize it as before. Calculate the Euclidean distance between the observations:
# Load the Boston dataset
data("Boston")
# Scale the dataset
boston_scaled <- scale(Boston) %>% as.data.frame()
# Calculate the Euclidean distance between the pairs of observations
dist_eu <- dist(boston_scaled)
summary(dist_eu)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4620 4.8240 4.9110 6.1860 14.4000
Run the k-means algorithm with 3 clusters and visualize the results:
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
Calculate te total of within cluster sum of squares (TWCSS) when the number of cluster changes from 1 to 10.
# Set seed to get reproducible results
set.seed(123)
# Determine the number of clusters
k_max <- 10
# Calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
The optimal number of clusters is when the total WCSS drops radically so based on the graph 2 seems to be the optimal number of clusters. Perform k-means with 2 clusters and visualize the results.
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 2)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
Perform k-means clustering with 3 clusters on the scaled Boston datset. Use the cluster asigments as the target variable for LDA analysis.
# Set seed to get reproducible results
set.seed(123)
# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# Add the cluster assingment to the dataset
boston_scaled$kmeans_cluster <- km$cluster
# linear discriminant analysis
lda.fit <- lda(kmeans_cluster ~ ., data = boston_scaled)
The LDA biplot:
# plot the lda results
plot(lda.fit, dimen = 2, col=boston_scaled$kmeans_cluster,
pch=boston_scaled$kmeans_cluster)
lda.arrows(lda.fit, myscale = 2)
Based on the biplot the most influencal linear separators are:
# Load libraries
library(corrplot)
library(dplyr)
library(tidyr)
Load the dataset from file.
human <- read.csv('data/human.csv', row.names = 1)
head(human)
## Edu2.FM Labo.FM Edu.Exp Life.Exp GNI Mat.Mor Ado.Birth
## Norway 1.0072389 0.8908297 17.5 81.6 64992 4 7.8
## Australia 0.9968288 0.8189415 20.2 82.4 42261 6 12.1
## Switzerland 0.9834369 0.8251001 15.8 83.0 56431 6 1.9
## Denmark 0.9886128 0.8840361 18.7 80.2 44025 5 5.1
## Netherlands 0.9690608 0.8286119 17.9 81.6 45435 6 6.2
## Germany 0.9927835 0.8072289 16.5 80.9 43919 7 3.8
## Parli.F
## Norway 39.6
## Australia 30.5
## Switzerland 28.5
## Denmark 38.0
## Netherlands 36.9
## Germany 36.9
str(human)
## 'data.frame': 155 obs. of 8 variables:
## $ Edu2.FM : num 1.007 0.997 0.983 0.989 0.969 ...
## $ Labo.FM : num 0.891 0.819 0.825 0.884 0.829 ...
## $ Edu.Exp : num 17.5 20.2 15.8 18.7 17.9 16.5 18.6 16.5 15.9 19.2 ...
## $ Life.Exp : num 81.6 82.4 83 80.2 81.6 80.9 80.9 79.1 82 81.8 ...
## $ GNI : int 64992 42261 56431 44025 45435 43919 39568 52947 42155 32689 ...
## $ Mat.Mor : int 4 6 6 5 6 7 9 28 11 8 ...
## $ Ado.Birth: num 7.8 12.1 1.9 5.1 6.2 3.8 8.2 31 14.5 25.3 ...
## $ Parli.F : num 39.6 30.5 28.5 38 36.9 36.9 19.9 19.4 28.2 31.4 ...
The dataset contains 155 observation of 8 variables. The dataset combines several indicators from most countries in the world. The countries are the rownames of the data.frame and the variables are:
Health and knowledge
Empowerment
Visualize the distribution of the variabels and their depencies
library(GGally)
library(ggplot2)
# Define the plot.
p <- ggpairs(human, mapping = aes(alpha=0.3),
lower = list(combo = wrap("facethist", bins = 20)))
# Draw the plot
p
Figure: Pairs-plot of the variables in the dataset.
Most of the variables have very skewed distributions such Lif.Exp and GNI. Only Edu.Exp seems to be almost normally distributed.
Next visualize the correlation between the variables
cor(human) %>% corrplot()
Figure: Correlation of the variables in the dataset.
There is a group of highly correlated variables: Edu.Exp, Lif.Exp, GNI, Mat.Mor, Ado.Birth. For example Edu.Exp and Lif.Exp are positively correlated and Edu.Exp and Mat.Mor are negatively correlated.
Summaries of the variables
summary(human)
## Edu2.FM Labo.FM Edu.Exp Life.Exp
## Min. :0.1717 Min. :0.1857 Min. : 5.40 Min. :49.00
## 1st Qu.:0.7264 1st Qu.:0.5984 1st Qu.:11.25 1st Qu.:66.30
## Median :0.9375 Median :0.7535 Median :13.50 Median :74.20
## Mean :0.8529 Mean :0.7074 Mean :13.18 Mean :71.65
## 3rd Qu.:0.9968 3rd Qu.:0.8535 3rd Qu.:15.20 3rd Qu.:77.25
## Max. :1.4967 Max. :1.0380 Max. :20.20 Max. :83.50
## GNI Mat.Mor Ado.Birth Parli.F
## Min. : 581 Min. : 1.0 Min. : 0.60 Min. : 0.00
## 1st Qu.: 4198 1st Qu.: 11.5 1st Qu.: 12.65 1st Qu.:12.40
## Median : 12040 Median : 49.0 Median : 33.60 Median :19.30
## Mean : 17628 Mean : 149.1 Mean : 47.16 Mean :20.91
## 3rd Qu.: 24512 3rd Qu.: 190.0 3rd Qu.: 71.95 3rd Qu.:27.95
## Max. :123124 Max. :1100.0 Max. :204.80 Max. :57.50
Perform principal component analysis (PCA) using singular value decomposition (SVD) method for un-standardized dataset. The variability captured by the principal components:
# perform principal component analysis (with the SVD method)
pca_human <- prcomp(human)
# create and print out a summary of pca_human
s <- summary(pca_human)
s
## Importance of components:
## PC1 PC2 PC3 PC4 PC5 PC6 PC7
## Standard deviation 1.854e+04 185.5219 25.19 11.45 3.766 1.566 0.1912
## Proportion of Variance 9.999e-01 0.0001 0.00 0.00 0.000 0.000 0.0000
## Cumulative Proportion 9.999e-01 1.0000 1.00 1.00 1.000 1.000 1.0000
## PC8
## Standard deviation 0.1591
## Proportion of Variance 0.0000
## Cumulative Proportion 1.0000
# rounded percentages of variance captured by each PC
pca_pr <- round(100*s$importance[2,], digits = 1)
# create object pc_lab to be used as axis labels
pc_lab <- paste0(names(pca_pr), " (", pca_pr, "%)")
# draw a biplot of the principal component representation and the original variables
biplot(pca_human, choices = 1:2, cex=c(0.7,1), col=c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
Figure: Countries plotted againts the two first principal components for the un-standardized dataset. Gross national income per capita (GNI) seems to explain most of the variation in the dataset. This follows from the fact that the values of the GNI variable are much larger than the other variables and that is why it dominates in the PCA analysis of the un-standardized dataset.
Next, standardize the dataset so that every variable has a mean of 0 and a standard deviation of 1 and perform PCA on the standardized dataset. The summary for the standardized variables
# standardize the variables
human_std <- scale(human)
# print out summaries of the standardized variables
summary(human_std)
## Edu2.FM Labo.FM Edu.Exp Life.Exp
## Min. :-2.8189 Min. :-2.6247 Min. :-2.7378 Min. :-2.7188
## 1st Qu.:-0.5233 1st Qu.:-0.5484 1st Qu.:-0.6782 1st Qu.:-0.6425
## Median : 0.3503 Median : 0.2316 Median : 0.1140 Median : 0.3056
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5958 3rd Qu.: 0.7350 3rd Qu.: 0.7126 3rd Qu.: 0.6717
## Max. : 2.6646 Max. : 1.6632 Max. : 2.4730 Max. : 1.4218
## GNI Mat.Mor Ado.Birth Parli.F
## Min. :-0.9193 Min. :-0.6992 Min. :-1.1325 Min. :-1.8203
## 1st Qu.:-0.7243 1st Qu.:-0.6496 1st Qu.:-0.8394 1st Qu.:-0.7409
## Median :-0.3013 Median :-0.4726 Median :-0.3298 Median :-0.1403
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.3712 3rd Qu.: 0.1932 3rd Qu.: 0.6030 3rd Qu.: 0.6127
## Max. : 5.6890 Max. : 4.4899 Max. : 3.8344 Max. : 3.1850
The variability captured by the principal components
# perform principal component analysis (with the SVD method)
pca_human_std <- prcomp(human_std)
# create and print out a summary of pca_human
s <- summary(pca_human_std)
s
## Importance of components:
## PC1 PC2 PC3 PC4 PC5 PC6
## Standard deviation 2.0708 1.1397 0.87505 0.77886 0.66196 0.53631
## Proportion of Variance 0.5361 0.1624 0.09571 0.07583 0.05477 0.03595
## Cumulative Proportion 0.5361 0.6984 0.79413 0.86996 0.92473 0.96069
## PC7 PC8
## Standard deviation 0.45900 0.32224
## Proportion of Variance 0.02634 0.01298
## Cumulative Proportion 0.98702 1.00000
# rounded percentages of variance captured by each PC
pca_pr <- round(100*s$importance[2,], digits = 1)
# create object pc_lab to be used as axis labels
pc_lab <- paste0(names(pca_pr), " (", pca_pr, "%)")
# draw a biplot of the principal component representation and the original variables
biplot(pca_human_std, choices = 1:2, cex=c(0.7,1), col=c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
Figure 4. Countries plotted againts the two first principal components for the standardized dataset.
The expected years of education, life expectancy at birth, gross national income per capita and the ratio of women to men with at least secondary education seem to correlate with each other as well as PC1. Maternal mortality and adolescent birth rates are inversly correlated to the former variables and also correlated with PC1. The ratio of women to men in the labor force is correlated with the percentage of women representatives in the parliament and this is correlated with PC2.
The first pricipal component seems to separate the countries based on variables related to health, education and wealth. The second pricipal component captures the variability in the participation of women to work and polital life in the society.
Load the ‘tea’ dataset from the package FactoMineR. The dataset represents a questionnaire on tea made in 300 inviduals: how they drink tea (18 questions), what are their product’s perception (12) and some personal details. The structure of the dataset:
library("FactoMineR")
data(tea)
str(tea)
## 'data.frame': 300 obs. of 36 variables:
## $ breakfast : Factor w/ 2 levels "breakfast","Not.breakfast": 1 1 2 2 1 2 1 2 1 1 ...
## $ tea.time : Factor w/ 2 levels "Not.tea time",..: 1 1 2 1 1 1 2 2 2 1 ...
## $ evening : Factor w/ 2 levels "evening","Not.evening": 2 2 1 2 1 2 2 1 2 1 ...
## $ lunch : Factor w/ 2 levels "lunch","Not.lunch": 2 2 2 2 2 2 2 2 2 2 ...
## $ dinner : Factor w/ 2 levels "dinner","Not.dinner": 2 2 1 1 2 1 2 2 2 2 ...
## $ always : Factor w/ 2 levels "always","Not.always": 2 2 2 2 1 2 2 2 2 2 ...
## $ home : Factor w/ 2 levels "home","Not.home": 1 1 1 1 1 1 1 1 1 1 ...
## $ work : Factor w/ 2 levels "Not.work","work": 1 1 2 1 1 1 1 1 1 1 ...
## $ tearoom : Factor w/ 2 levels "Not.tearoom",..: 1 1 1 1 1 1 1 1 1 2 ...
## $ friends : Factor w/ 2 levels "friends","Not.friends": 2 2 1 2 2 2 1 2 2 2 ...
## $ resto : Factor w/ 2 levels "Not.resto","resto": 1 1 2 1 1 1 1 1 1 1 ...
## $ pub : Factor w/ 2 levels "Not.pub","pub": 1 1 1 1 1 1 1 1 1 1 ...
## $ Tea : Factor w/ 3 levels "black","Earl Grey",..: 1 1 2 2 2 2 2 1 2 1 ...
## $ How : Factor w/ 4 levels "alone","lemon",..: 1 3 1 1 1 1 1 3 3 1 ...
## $ sugar : Factor w/ 2 levels "No.sugar","sugar": 2 1 1 2 1 1 1 1 1 1 ...
## $ how : Factor w/ 3 levels "tea bag","tea bag+unpackaged",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ where : Factor w/ 3 levels "chain store",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ price : Factor w/ 6 levels "p_branded","p_cheap",..: 4 6 6 6 6 3 6 6 5 5 ...
## $ age : int 39 45 47 23 48 21 37 36 40 37 ...
## $ sex : Factor w/ 2 levels "F","M": 2 1 1 2 2 2 2 1 2 2 ...
## $ SPC : Factor w/ 7 levels "employee","middle",..: 2 2 4 6 1 6 5 2 5 5 ...
## $ Sport : Factor w/ 2 levels "Not.sportsman",..: 2 2 2 1 2 2 2 2 2 1 ...
## $ age_Q : Factor w/ 5 levels "15-24","25-34",..: 3 4 4 1 4 1 3 3 3 3 ...
## $ frequency : Factor w/ 4 levels "1/day","1 to 2/week",..: 1 1 3 1 3 1 4 2 3 3 ...
## $ escape.exoticism: Factor w/ 2 levels "escape-exoticism",..: 2 1 2 1 1 2 2 2 2 2 ...
## $ spirituality : Factor w/ 2 levels "Not.spirituality",..: 1 1 1 2 2 1 1 1 1 1 ...
## $ healthy : Factor w/ 2 levels "healthy","Not.healthy": 1 1 1 1 2 1 1 1 2 1 ...
## $ diuretic : Factor w/ 2 levels "diuretic","Not.diuretic": 2 1 1 2 1 2 2 2 2 1 ...
## $ friendliness : Factor w/ 2 levels "friendliness",..: 2 2 1 2 1 2 2 1 2 1 ...
## $ iron.absorption : Factor w/ 2 levels "iron absorption",..: 2 2 2 2 2 2 2 2 2 2 ...
## $ feminine : Factor w/ 2 levels "feminine","Not.feminine": 2 2 2 2 2 2 2 1 2 2 ...
## $ sophisticated : Factor w/ 2 levels "Not.sophisticated",..: 1 1 1 2 1 1 1 2 2 1 ...
## $ slimming : Factor w/ 2 levels "No.slimming",..: 1 1 1 1 1 1 1 1 1 1 ...
## $ exciting : Factor w/ 2 levels "exciting","No.exciting": 2 1 2 2 2 2 2 2 2 2 ...
## $ relaxing : Factor w/ 2 levels "No.relaxing",..: 1 1 2 2 2 2 2 2 2 2 ...
## $ effect.on.health: Factor w/ 2 levels "effect on health",..: 2 2 2 2 2 2 2 2 2 2 ...
The dataset contains 300 observations of 36 variables. Visualize the dataset:
# visualize the dataset
gather(tea) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar() + theme(axis.text.x = element_text(angle = 45, hjust = 1, size = 10))
Figure: Variables and the distribution of their values in the ‘tea’ dataset.
Perform multiple correspondance analysis (MCA) on the dataset using a subset of variables: Tea, How, how, sugar, where, lunch:
# Column names to keep in the dataset
keep_columns <- c("Tea", "How", "how", "sugar", "where", "lunch")
# Select the 'keep_columns' to create a new dataset
tea_time <- dplyr::select(tea, one_of(keep_columns))
# Multiple correspondence analysis
mca <- MCA(tea_time, graph = FALSE)
# Summary of the model
summary(mca)
##
## Call:
## MCA(X = tea_time, graph = FALSE)
##
##
## Eigenvalues
## Dim.1 Dim.2 Dim.3 Dim.4 Dim.5 Dim.6
## Variance 0.279 0.261 0.219 0.189 0.177 0.156
## % of var. 15.238 14.232 11.964 10.333 9.667 8.519
## Cumulative % of var. 15.238 29.471 41.435 51.768 61.434 69.953
## Dim.7 Dim.8 Dim.9 Dim.10 Dim.11
## Variance 0.144 0.141 0.117 0.087 0.062
## % of var. 7.841 7.705 6.392 4.724 3.385
## Cumulative % of var. 77.794 85.500 91.891 96.615 100.000
##
## Individuals (the 10 first)
## Dim.1 ctr cos2 Dim.2 ctr cos2 Dim.3
## 1 | -0.298 0.106 0.086 | -0.328 0.137 0.105 | -0.327
## 2 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 3 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 4 | -0.530 0.335 0.460 | -0.318 0.129 0.166 | 0.211
## 5 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 6 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 7 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 8 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 9 | 0.143 0.024 0.012 | 0.871 0.969 0.435 | -0.067
## 10 | 0.476 0.271 0.140 | 0.687 0.604 0.291 | -0.650
## ctr cos2
## 1 0.163 0.104 |
## 2 0.735 0.314 |
## 3 0.062 0.069 |
## 4 0.068 0.073 |
## 5 0.062 0.069 |
## 6 0.062 0.069 |
## 7 0.062 0.069 |
## 8 0.735 0.314 |
## 9 0.007 0.003 |
## 10 0.643 0.261 |
##
## Categories (the 10 first)
## Dim.1 ctr cos2 v.test Dim.2 ctr
## black | 0.473 3.288 0.073 4.677 | 0.094 0.139
## Earl Grey | -0.264 2.680 0.126 -6.137 | 0.123 0.626
## green | 0.486 1.547 0.029 2.952 | -0.933 6.111
## alone | -0.018 0.012 0.001 -0.418 | -0.262 2.841
## lemon | 0.669 2.938 0.055 4.068 | 0.531 1.979
## milk | -0.337 1.420 0.030 -3.002 | 0.272 0.990
## other | 0.288 0.148 0.003 0.876 | 1.820 6.347
## tea bag | -0.608 12.499 0.483 -12.023 | -0.351 4.459
## tea bag+unpackaged | 0.350 2.289 0.056 4.088 | 1.024 20.968
## unpackaged | 1.958 27.432 0.523 12.499 | -1.015 7.898
## cos2 v.test Dim.3 ctr cos2 v.test
## black 0.003 0.929 | -1.081 21.888 0.382 -10.692 |
## Earl Grey 0.027 2.867 | 0.433 9.160 0.338 10.053 |
## green 0.107 -5.669 | -0.108 0.098 0.001 -0.659 |
## alone 0.127 -6.164 | -0.113 0.627 0.024 -2.655 |
## lemon 0.035 3.226 | 1.329 14.771 0.218 8.081 |
## milk 0.020 2.422 | 0.013 0.003 0.000 0.116 |
## other 0.102 5.534 | -2.524 14.526 0.197 -7.676 |
## tea bag 0.161 -6.941 | -0.065 0.183 0.006 -1.287 |
## tea bag+unpackaged 0.478 11.956 | 0.019 0.009 0.000 0.226 |
## unpackaged 0.141 -6.482 | 0.257 0.602 0.009 1.640 |
##
## Categorical variables (eta2)
## Dim.1 Dim.2 Dim.3
## Tea | 0.126 0.108 0.410 |
## How | 0.076 0.190 0.394 |
## how | 0.708 0.522 0.010 |
## sugar | 0.065 0.001 0.336 |
## where | 0.702 0.681 0.055 |
## lunch | 0.000 0.064 0.111 |
The first dimension explains 15% of the variation and the second dimension 14% variation in the data. Variables how and where have the strongest link to the first and second dimension out of all the analyzed variables.
Visualize the MCA results:
# Visualize MCA
plot(mca, invisible=c("ind"), habillage= "quali")
Figure: Variable biplot of the MCA analysis results on the ‘tea’ dataset with variables Tea, How, how, sugar, where, lunch.
Based on the plot individuals who use unpackaged tea also tend to buy their tea from the tea shops and prefer green tea. On the other individuals who use tea bags buy them often from chain stores.
# Load libraries
library(dplyr)
library(tidyr)
library(ggplot2)
In the BPRS dataset 40 male subjects were randomly assigned to one of two treatment groups and each subject was rated on the brief psychiatric rating scale (BPRS) measured before treatment began (week 0) and then at weekly intervals for eight weeks. The BPRS assesses the level of 18 symptom constructs such as hostility, suspiciousness, hallucinations and grandiosity; each of these is rated from one (not present) to seven (extremely severe). The scale is used to evaluate patients suspected of having schizophrenia.
Prepare the data set:
# Prepare the data
source("data/meet_and_repeat.R")
# Structure of the data
str(BPRSL)
## 'data.frame': 360 obs. of 5 variables:
## $ treatment: Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 1 1 ...
## $ subject : Factor w/ 20 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ weeks : chr "week0" "week0" "week0" "week0" ...
## $ bprs : int 42 58 54 55 72 48 71 30 41 57 ...
## $ week : int 0 0 0 0 0 0 0 0 0 0 ...
Plot the bprs values over time for each 40 individuals by treatment group:
# Draw the plot
ggplot(BPRSL, aes(x = week, y = bprs, linetype = subject)) +
geom_line() +
scale_linetype_manual(values = rep(1:10, times=4)) +
facet_grid(. ~ treatment, labeller = label_both) +
theme(legend.position = "none") +
scale_y_continuous(limits = c(min(BPRSL$bprs), max(BPRSL$bprs)))
The BPRS score seems to decrease over time as well as the variability of the score in both treatment groups.
Next, standaridize the scores for each time point by substracting the mean bprs for a given time point for all the values at that time point and dividing by the standard deviation of bprs at that time point.
# Standardise the variable bprs
BPRSL <- BPRSL %>%
group_by(week) %>%
mutate(stdbprs = scale(bprs)) %>%
ungroup()
# Glimpse the data
glimpse(BPRSL)
## Observations: 360
## Variables: 6
## $ treatment <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
## $ subject <fct> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1...
## $ weeks <chr> "week0", "week0", "week0", "week0", "week0", "week0"...
## $ bprs <int> 42, 58, 54, 55, 72, 48, 71, 30, 41, 57, 30, 55, 36, ...
## $ week <int> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0...
## $ stdbprs <dbl> -0.4245908, 0.7076513, 0.4245908, 0.4953559, 1.69836...
Plot standardized bprs values:
# Plot again with the standardised bprs
ggplot(BPRSL, aes(x = week, y = stdbprs, linetype = subject)) +
geom_line() +
scale_linetype_manual(values = rep(1:10, times=4)) +
facet_grid(. ~ treatment, labeller = label_both) +
scale_y_continuous(name = "standardized bprs")
Next plot the mean bprs for each time point for the two different treatment groups and also add the standard error of the mean to the plots:
\[se = \frac{sd(x)}{\sqrt{n}}\]
# Mean profiles
################
# Number of weeks, baseline (week 0) included
n <- BPRSL$week %>% unique() %>% length()
# Summary data with mean and standard error of bprs by treatment and week
BPRSS <- BPRSL %>%
group_by(treatment, week) %>%
summarise(mean = mean(bprs), se = sd(bprs)/sqrt(n) ) %>%
ungroup()
# Glimpse the data
glimpse(BPRSS)
## Observations: 18
## Variables: 4
## $ treatment <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2
## $ week <int> 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8
## $ mean <dbl> 47.00, 46.80, 43.55, 40.90, 36.60, 32.70, 29.70, 29....
## $ se <dbl> 4.534468, 5.173708, 4.003617, 3.744626, 3.259534, 2....
# Plot the mean profiles
ggplot(BPRSS, aes(x = week, y = mean, linetype = treatment, shape = treatment)) +
geom_line() +
scale_linetype_manual(values = c(1,2)) +
geom_point(size=3) +
scale_shape_manual(values = c(1,2)) +
geom_errorbar(aes(ymin=mean-se, ymax=mean+se, linetype="1"), width=0.3) +
theme(legend.position = c(0.8,0.8)) +
scale_y_continuous(name = "mean(bprs) +/- se(bprs)")
The mean profiles seem to overlap completly when taking the strandard errors of the mean estimates into account. This suggests there is only a small difference between the treatment groups when looking at the mean profiles.
Compare the mean bprs values between the treatment groups on weeks 1-8 by plotting the distribution of the mean bprs values for the two groups:
# Create a summary data by treatment and subject with mean as the summary variable (ignoring baseline week 0).
BPRSL8S <- BPRSL %>%
filter(week > 0) %>%
group_by(treatment, subject) %>%
summarise(mean=mean(bprs)) %>%
ungroup()
# Glimpse the data
glimpse(BPRSL8S)
## Observations: 40
## Variables: 3
## $ treatment <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
## $ subject <fct> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1...
## $ mean <dbl> 41.500, 43.125, 35.375, 52.625, 50.375, 34.000, 37.1...
# Draw a boxplot of the mean versus treatment
ggplot(BPRSL8S, aes(x = treatment, y = mean)) +
geom_boxplot() +
stat_summary(fun.y = "mean", geom = "point", shape=23, size=4, fill = "white") +
scale_y_continuous(name = "mean(bprs), weeks 1-8")
There seems to be an outlier in group 2 with a mean bprs value over 70. This might bias the results so remove it and plot again:
BPRSL8S1 <- BPRSL %>%
filter(week > 0) %>%
group_by(treatment, subject) %>%
summarise(mean=mean(bprs)) %>%
ungroup() %>%
filter(mean < 70)
# Draw a boxplot of the mean versus treatment
ggplot(BPRSL8S1, aes(x = treatment, y = mean)) +
geom_boxplot() +
stat_summary(fun.y = "mean", geom = "point", shape=23, size=4, fill = "white") +
scale_y_continuous(name = "mean(bprs), weeks 1-8")
Based on the plot it looks like the mean brps is a bit lower in the treatment group 2 but the difference is small compared to the variation of the mean bprs inside the treatment groups.
Let’s perform a t-test comparing the mean bprs values between the treatment groups:
# Perform a two-sample t-test
t.test(mean ~ treatment, data = BPRSL8S1, var.equal = TRUE)
##
## Two Sample t-test
##
## data: mean by treatment
## t = 0.52095, df = 37, p-value = 0.6055
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -4.232480 7.162085
## sample estimates:
## mean in group 1 mean in group 2
## 36.16875 34.70395
There is no statistically significant difference between the groups which is indicated also by the 95% confidence interval which includes the zero.
The baseline bprs value might be correlated with the chosen summary measure. Let’s add that to our model to see if that will affect the difference between the treatment groups:
# Add the baseline from the original data as a new variable to the summary data
BPRSL8S2 <- BPRSL8S %>%
mutate(baseline = BPRS$week0)
# Fit the linear model with the mean as the response
fit <- lm(mean ~ treatment + baseline, data = BPRSL8S2)
# Compute the analysis of variance table for the fitted model with anova()
anova(fit)
## Analysis of Variance Table
##
## Response: mean
## Df Sum Sq Mean Sq F value Pr(>F)
## treatment 1 1.55 1.55 0.025 0.8752
## baseline 1 1869.97 1869.97 30.174 3.051e-06 ***
## Residuals 37 2292.97 61.97
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The baseline bprs values is strongly associated to the bprs values taken after treatment has begun, but there is still no evidence of a treatment difference even after conditioning on the baseline value.
The RATS dataset comes from a nutrition study conducted in three groups of rats. The groups were put on different diets, and each animal’s body weight (grams) was recorded repeatedly (approximately) weekly, except in week seven when two recordings were taken) over a 9-week period.
str(RATSL)
## 'data.frame': 176 obs. of 5 variables:
## $ ID : Factor w/ 16 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ Group : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 1 1 1 2 2 ...
## $ WD : chr "WD1" "WD1" "WD1" "WD1" ...
## $ Weight: int 240 225 245 260 255 260 275 245 410 405 ...
## $ Time : int 1 1 1 1 1 1 1 1 1 1 ...
Plot the RATSL dataset:
# Plot the RATSL data
ggplot(RATSL, aes(x = Time, y = Weight, group = ID)) +
geom_line(aes(linetype = Group)) +
scale_x_continuous(name = "Time (days)", breaks = seq(0, 60, 10)) +
scale_y_continuous(name = "Weight (grams)") +
theme(legend.position = "top")
The weight of the rats in group 1 is lower at the start of the follow-up compared to the rats in groups 2 and 3 and stays lower during the follow-up.
Let’s fit a linear regression model where Weight is the outcome variable and Group and Time are the explanatory variables:
\[Weight \sim Group + Time\].
Here we are assuming the repeated measures of the same animal to be independent which is highly unlikely.
# create a regression model RATS_reg
RATS_reg <- lm(Weight ~ Time + Group, data=RATSL)
# print out a summary of the model
summary(RATS_reg)
##
## Call:
## lm(formula = Weight ~ Time + Group, data = RATSL)
##
## Residuals:
## Min 1Q Median 3Q Max
## -60.643 -24.017 0.697 10.837 125.459
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 244.0689 5.7725 42.281 < 2e-16 ***
## Time 0.5857 0.1331 4.402 1.88e-05 ***
## Group2 220.9886 6.3402 34.855 < 2e-16 ***
## Group3 262.0795 6.3402 41.336 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 34.34 on 172 degrees of freedom
## Multiple R-squared: 0.9283, Adjusted R-squared: 0.9271
## F-statistic: 742.6 on 3 and 172 DF, p-value: < 2.2e-16
From the summary of the model we can see that weight is statistically significantly higher in groups 2 and 3 compared to group 1. Also, the regression coeffiecient of time is smaller than one and statistically siginifcant so the weight of the animals seems to go down on average during the follow-up.
Let’s fit a random intercept model using the same two explanatory variables: Time and Group. To allow the rats to have a different weight at the start of the follow-up we use the identiy of the rats as the random effect. Fitting a random intercept model allows the linear regression fit for each rat to differ in intercept from other rats.
# access library lme4
library(lme4)
# Create a random intercept model
RATS_ref <- lmer(Weight ~ Time + Group + (1 | ID), data = RATSL, REML = FALSE)
# Print the summary of the model
summary(RATS_ref)
## Linear mixed model fit by maximum likelihood ['lmerMod']
## Formula: Weight ~ Time + Group + (1 | ID)
## Data: RATSL
##
## AIC BIC logLik deviance df.resid
## 1333.2 1352.2 -660.6 1321.2 170
##
## Scaled residuals:
## Min 1Q Median 3Q Max
## -3.5386 -0.5581 -0.0494 0.5693 3.0990
##
## Random effects:
## Groups Name Variance Std.Dev.
## ID (Intercept) 1085.92 32.953
## Residual 66.44 8.151
## Number of obs: 176, groups: ID, 16
##
## Fixed effects:
## Estimate Std. Error t value
## (Intercept) 244.06890 11.73107 20.80
## Time 0.58568 0.03158 18.54
## Group2 220.98864 20.23577 10.92
## Group3 262.07955 20.23577 12.95
##
## Correlation of Fixed Effects:
## (Intr) Time Group2
## Time -0.090
## Group2 -0.575 0.000
## Group3 -0.575 0.000 0.333
From the summary of the model we can see that even after allowing different weight of the rats at start of the follow-up the rats are heawier in groups 2 and 3 compared to group 1 and the weight of the rats seems to decrease during the follow-up.
Next, add random slope to the model of the rat growth data. Using a random intercept and random slope model allows the linear regression fits for each rat to differ in intercept and slope. This way it is possible to take into account that the rats start with different weights and their weights might change overtime with different rates but also to analyse the effect of time in general to the change of weight.
# create a random intercept and random slope model
RATS_ref1 <- lmer(Weight ~ Time + Group + (Time | ID), data = RATSL, REML = FALSE)
# print a summary of the model
summary(RATS_ref1)
## Linear mixed model fit by maximum likelihood ['lmerMod']
## Formula: Weight ~ Time + Group + (Time | ID)
## Data: RATSL
##
## AIC BIC logLik deviance df.resid
## 1194.2 1219.6 -589.1 1178.2 168
##
## Scaled residuals:
## Min 1Q Median 3Q Max
## -3.2261 -0.4322 0.0555 0.5638 2.8827
##
## Random effects:
## Groups Name Variance Std.Dev. Corr
## ID (Intercept) 1140.5363 33.7718
## Time 0.1122 0.3349 -0.22
## Residual 19.7456 4.4436
## Number of obs: 176, groups: ID, 16
##
## Fixed effects:
## Estimate Std. Error t value
## (Intercept) 246.45727 11.81526 20.859
## Time 0.58568 0.08548 6.852
## Group2 214.58736 20.17983 10.634
## Group3 258.92732 20.17983 12.831
##
## Correlation of Fixed Effects:
## (Intr) Time Group2
## Time -0.166
## Group2 -0.569 0.000
## Group3 -0.569 0.000 0.333
From the summary of the model we see that the using the random intercept and random slope model the rats in groups 2 and 3 are heavier than rats in group 1 and that the weight of the rats seems to go down on average.
Compare the random intercept and random intercept and slope models by performing likelihood ratio test:
# perform an ANOVA test on the two models
anova(RATS_ref1, RATS_ref)
## Data: RATSL
## Models:
## RATS_ref: Weight ~ Time + Group + (1 | ID)
## RATS_ref1: Weight ~ Time + Group + (Time | ID)
## Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
## RATS_ref 6 1333.2 1352.2 -660.58 1321.2
## RATS_ref1 8 1194.2 1219.6 -589.11 1178.2 142.94 2 < 2.2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The p-value is highly significant and the log-likelihood of the random intercept and random slope is greates than that of the random slope model indicating that it fits the data better (the fit is better the closer to 0 the log-likelihood of the model is).
To test if the growth profiles of the rats differ between the groups we will fit a random intercept and slope model that allows for a Group × Time interaction.
# create a random intercept and random slope model with the interaction
RATS_ref2 <- lmer(Weight ~ Time + Group + Time * Group + (Time | ID), data = RATSL, REML = FALSE)
# print a summary of the model
summary(RATS_ref2)
## Linear mixed model fit by maximum likelihood ['lmerMod']
## Formula: Weight ~ Time + Group + Time * Group + (Time | ID)
## Data: RATSL
##
## AIC BIC logLik deviance df.resid
## 1185.9 1217.6 -582.9 1165.9 166
##
## Scaled residuals:
## Min 1Q Median 3Q Max
## -3.2669 -0.4249 0.0726 0.6034 2.7513
##
## Random effects:
## Groups Name Variance Std.Dev. Corr
## ID (Intercept) 1.107e+03 33.2763
## Time 4.925e-02 0.2219 -0.15
## Residual 1.975e+01 4.4436
## Number of obs: 176, groups: ID, 16
##
## Fixed effects:
## Estimate Std. Error t value
## (Intercept) 251.65165 11.80279 21.321
## Time 0.35964 0.08215 4.378
## Group2 200.66549 20.44303 9.816
## Group3 252.07168 20.44303 12.330
## Time:Group2 0.60584 0.14229 4.258
## Time:Group3 0.29834 0.14229 2.097
##
## Correlation of Fixed Effects:
## (Intr) Time Group2 Group3 Tm:Gr2
## Time -0.160
## Group2 -0.577 0.092
## Group3 -0.577 0.092 0.333
## Time:Group2 0.092 -0.577 -0.160 -0.053
## Time:Group3 0.092 -0.577 -0.053 -0.160 0.333
From the summary we can see that the interaction of time and weight is stronger in groups 2 and 3 compared to group 1: the rats gain weight faster in groups 2 and 3 compared to group 1.
Compare the random intercept and random slope model to the random intercept and random slope with Time and Weight interaction using anova:
# perform an ANOVA test on the two models
anova(RATS_ref2, RATS_ref1)
## Data: RATSL
## Models:
## RATS_ref1: Weight ~ Time + Group + (Time | ID)
## RATS_ref2: Weight ~ Time + Group + Time * Group + (Time | ID)
## Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
## RATS_ref1 8 1194.2 1219.6 -589.11 1178.2
## RATS_ref2 10 1185.9 1217.6 -582.93 1165.9 12.361 2 0.00207 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The model with Time and Weight interaction seems to fit the data better based on the smaller log-likelihood of the model. And the difference is statistically significant.
Finally, visualize the observed weights and the fitted values for weight from the last model:
# draw the plot of RATSL with the observed Weight values
ggplot(RATSL, aes(x = Time, y = Weight, group = ID)) +
geom_line(aes(linetype = Group)) +
scale_x_continuous(name = "Time (days)", breaks = seq(0, 60, 20)) +
scale_y_continuous(name = "Observed weight (grams)") +
theme(legend.position = "top")
# Create a vector of the fitted values
Fitted <- fitted(RATS_ref2)
# Create a new column fitted to RATSL
RATSL <- mutate(RATSL, Fitted=Fitted)
# draw the plot of RATSL with the Fitted values of weight
ggplot(RATSL, aes(x = Time, y = Fitted, group = ID)) +
geom_line(aes(linetype = Group)) +
scale_x_continuous(name = "Time (days)", breaks = seq(0, 60, 20)) +
scale_y_continuous(name = "Fitted weight (grams)") +
theme(legend.position = "top")